Question-142333

Question Number 142333 by mohammad17 last updated on 30/May/21

Commented by mohammad17 last updated on 30/May/21

$${please}\:{sir}\:{help}\:{me} \\ $$

Answered by Dwaipayan Shikari last updated on 30/May/21

lim_(x→0) (((x+4)^(3/2) −8)/x)=((8((1+(x/4))^(3/2) −1))/x)=8((1+(x/4).(3/2)−1)/x)=3

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+\mathrm{4}\right)^{\mathrm{3}/\mathrm{2}} −\mathrm{8}}{{x}}=\frac{\mathrm{8}\left(\left(\mathrm{1}+\frac{{x}}{\mathrm{4}}\right)^{\mathrm{3}/\mathrm{2}} −\mathrm{1}\right)}{{x}}=\mathrm{8}\frac{\mathrm{1}+\frac{{x}}{\mathrm{4}}.\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}}{{x}}=\mathrm{3} \\ $$

Commented by mathmax by abdo last updated on 30/May/21

let f(x)=(((x+4)^(3/2) −8)/x) ⇒f(x)=((4^(3/2) (1+(x/4))^(3/2) −8)/x) =((8{(1+(x/4))^(3/2) −1})/x)∼((8(1+((3x)/8)−1))/x)=3 ⇒lim_(x→0) f(x)=3

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}+\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{8}}{\mathrm{x}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{8}}{\mathrm{x}} \\ $$$$=\frac{\mathrm{8}\left\{\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{1}\right\}}{\mathrm{x}}\sim\frac{\mathrm{8}\left(\mathrm{1}+\frac{\mathrm{3x}}{\mathrm{8}}−\mathrm{1}\right)}{\mathrm{x}}=\mathrm{3}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{3} \\ $$$$ \\ $$

Answered by bramlexs22 last updated on 30/May/21

f(x)=lim_(x→0) (((√((x+4)^3 ))−8)/x) f(x)=lim_(x→0) (((x+4)^3 −64)/x).lim_(x→0) (1/( (√((x+4)^3 ))+8)) = lim_(x→0) ((x(x^2 +12x+48))/x). (1/(16)) = ((48)/(16)) = 3

$${f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\left({x}+\mathrm{4}\right)^{\mathrm{3}} }−\mathrm{8}}{{x}} \\ $$$${f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+\mathrm{4}\right)^{\mathrm{3}} −\mathrm{64}}{{x}}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\left({x}+\mathrm{4}\right)^{\mathrm{3}} }+\mathrm{8}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{48}\right)}{{x}}.\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$=\:\frac{\mathrm{48}}{\mathrm{16}}\:=\:\mathrm{3}\: \\ $$

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